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Answer :

`16.38mL,34.18L`Solution :

Smallest volume of `AgNO_(3)` would be required when the entire mass is due to highest molecular weight constituent. <br> Hence, for smallest volume, the whole mass should be of `BaCl_(2). 2H_(2)O` <br> `m` mol of `BaCl_(2)2H_(2)O=(0.3)/(244)xx1000=1.229m` mol `m` mol of `AgNO_(3)`required `=2xx1.229 = 2.458` <br> Volume of `AgNO_(3)` required `=(2.458)/(0.15)=16.38mL` <br> Largest volume of `AgNO_(3)` would be required when entire mass is due to lowest molecular weight constitute, i.e., `NaCl`. <br> `m` mol of `NaCl = (0.3)/(58.5) xx1000 = 5.128 = m` mol of `AgNO_(3)` required <br> implies Volume of `AgNO_(3)` required `=(5.128)/(0.16) = 31.18ml. ("largest")`